在Java中创建文件时，如何确定允许的最大路径长度

在Java中创建文件时，如何确定允许的最大路径长度。 我使用的是Java 7，因此如果有帮助的话，我可以使用Java NIO2，但我如何确定文件系统上允许的文件最大长度，并确保我不尝试创建具有如此无效路径的文件

我希望在尝试创建之前缩短路径（子字符串），而不是在创建之后处理问题，特别是因为异常/错误消息可能因文件系统而异

我不想硬性规定每个平台的代码长度，因为Windows应用程序可以访问OSX文件系统ectera，所以这根本不起作用

它仅在尝试创建文件时失败 i、 e

给予

java.nio.file.FileSystemException: C:\User\Mesh\kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt: The filename, directory name, or volume label syntax is incorrect.

    at sun.nio.fs.WindowsException.translateToIOException(WindowsException.java:86)
    at sun.nio.fs.WindowsException.rethrowAsIOException(WindowsException.java:97)
    at sun.nio.fs.WindowsException.rethrowAsIOException(WindowsException.java:102)
    at sun.nio.fs.WindowsFileSystemProvider.newByteChannel(WindowsFileSystemProvider.java:229)
    at java.nio.file.Files.newByteChannel(Files.java:315)
    at java.nio.file.Files.createFile(Files.java:586)

与Java6方法比较

try
        {
            //File file = new File("C:/User/Mesh/kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt");
            Path p = Paths.get("C:/User/Mesh/kkkkkkkkkkkkkkkkkkkkkkkkkkrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.txt");
            p.toFile().createNewFile();
            //boolean result=file.isFile();
            //System.out.println("Result is:"+result);
        }
        catch(Exception e)
        {
            e.printStackTrace();
        }

给出了完全不同的错误

java.io.IOException: The filename, directory name, or volume label syntax is incorrect
    at java.io.WinNTFileSystem.createFileExclusively(Native Method)
    at java.io.File.createNewFile(File.java:947)

但实际上是不相关的，因为我想在创建文件之前做一些事情

编辑：

---

看起来255个字符是一个很好的默认值

尝试以下内容：

boolean success = true;
File testFile = new File("/a/really/long/name");
try {
    success = testFile.createNewFile();
} catch (IOException ioe) {
    // anyway check the exception, the problem could have been other
    // for example: duplicate name, lack of permissions, etc.
    success = false;
}

if (!success) {
    // error creating file
}

---

java.nio.file.FileSystem

中有一个方法

getPath

，可用于创建有效的路径名，否则会引发InvalidPathException。也许这会对您有所帮助。

这对它在任何系统上工作都有什么帮助呢？不仅仅是我自己的系统。要记住的一点是，如果您使用Eclipse，它有自己的（相当严重的）路径长度限制。我无法立即找到实现此值的方法。可能最好是在试图打开文件并在那里处理时捕获异常。我想要的是一种在创建文件之前检查文件名是否有效的方法。@PaulTaylor我认为这是不可能的。最简单的方法是尝试创建它，并在出现错误的情况下处理它，如上所示，这看起来很有用，但我认为现在我只需要将其编码为255个字符

boolean success = true;
File testFile = new File("/a/really/long/name");
try {
    success = testFile.createNewFile();
} catch (IOException ioe) {
    // anyway check the exception, the problem could have been other
    // for example: duplicate name, lack of permissions, etc.
    success = false;
}

if (!success) {
    // error creating file
}